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  4. A capacitor is connected to a 20 V battery through a resistance of 10 Ω. It is found that the potential difference across the capacitor rises to 2 V in 1 μ s. The capacitance of the capacitor is μ F . Given: ln ((10/9))=0.105

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Q. A capacitor is connected to a $20 \,V$ battery through a resistance of $10\, \Omega$. It is found that the potential difference across the capacitor rises to $2\, V$ in $1 \,\mu s$. The capacitance of the capacitor is ____ $\mu \,F .$
Given : $\ln \left(\frac{10}{9}\right)=0.105$

JEE MainJEE Main 2021Electrostatic Potential and Capacitance

Solution:

$V = V _{0}\left(1- e ^{- t / RC }\right)$
$2=20\left(1- e ^{- t / RC }\right)$
$\frac{1}{10}=1- e ^{- t / RC }$
$e ^{- t / RC }=\frac{9}{10}$
$e ^{ t / RC }=\frac{10}{9}$
$\frac{t}{R C}=\ell n\left(\frac{10}{9}\right) $
$\Rightarrow C=\frac{t}{R \ell n\left(\frac{10}{9}\right)}$
$C =\frac{10^{-6}}{10 \times .105}$
$=.95\, \mu F$